Question

lim n->infinit din suma din k=1 la n din 1/2^k + 1/3^k

Answer 1

$\displaystyle Vom~folosi~identitatea~\boxed{1+a+a^2+...+a^n= \frac{a^{n+1}-1}{a-1}}~. \\ \\ \sum\limits_{k=1}^n \left( \frac{1}{2^k}+ \frac{1}{3^k} \right)= \sum\limits_{k=1}^n \frac{1}{2^k}+ \sum\limits_{k=1}^n \frac{1}{3^k}. \\ \\ \sum\limits_{k=1}^n \frac{1}{2^k}= \frac{1}{2}+ \frac{1}{2^2}+ \frac{1}{2^3}+...+ \frac{1}{2^n}= \frac{1}{2} \left(1+ \frac{1}{2}+ \frac{1}{2^2}+...+ \frac{1}{2^{n-1}}\right)= \\ \\ = \frac{1}{2} \cdot \frac{\frac{1}{2^n}-1}{ \frac{1}{2}-1}= 1- \frac{1}{2^n}.$

$\displaystyle \sum\limits_{k=1}^n \frac{1}{3^k}= \frac{1}{3}+ \frac{1}{3^2}+ \frac{1}{3^3}+...+ \frac{1}{3^n}= \frac{1}{3} \left( 1+ \frac{1}{3}+ \frac{1}{3^2}+...+ \frac{1}{3^{n-1}}\right)= \\ \\ = \frac{1}{3} \cdot \frac{\frac{1}{3^n}-1}{\frac{1}{3}-1}= \frac{1- \frac{1}{3^n}}{2}.$

$\displaystyle \lim_{n \to \infty} \sum\limits_{k=1}^n \left( \frac{1}{2^k}+ \frac{1}{3^k} \right)= \lim_{n \to \infty} \left( 1-\frac{1}{2^n}+ \frac{1- \frac{1}{3^n}}{2} \right)= \\ \\ = \lim_{n \to \infty} \left( \frac{3}{2}- \frac{1}{2^n}- \frac{1}{2 \cdot 3^n} \right)= \frac{3}{2}.$